Calculate the super elevation to be provided on the horizontal curve of radius 100 m. Design speed is 50 km/h and the design coefficient of lateral friction of 0.15 is fully developed.
0.047
Super elevation, also known as banking of roads, is provided on horizontal curves to counteract the centrifugal force acting on vehicles, ensuring stability and preventing skidding. The amount of super elevation needed depends on factors like the design speed, radius of the curve, and the coefficient of lateral friction.
We are provided with the following information for the horizontal curve:
The standard formula for super elevation uses speed in meters per second (m/s). We need to convert the given speed from kilometers per hour (km/h) to m/s.
Conversion factor: \(1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18} \text{ m/s}\)
Therefore, Design speed \(V\) in m/s is:
\(V = 50 \text{ km/h} \times \frac{5}{18} \text{ m/s per km/h}\)
\(V = \frac{250}{18} \text{ m/s}\)
\(V \approx 13.889 \text{ m/s}\)
When both super elevation (e) and lateral friction (f) are considered for a design speed V and curve radius R, the relationship is given by the formula:
\(e + f = \frac{V^2}{gR}\)
Where:
Now, we substitute the converted speed and the given values into the formula:
\(e + 0.15 = \frac{(13.889 \text{ m/s})^2}{(9.81 \text{ m/s}^2) \times (100 \text{ m})}\)
\(e + 0.15 = \frac{192.994}{(9.81 \times 100)}\)
\(e + 0.15 = \frac{192.994}{981}\)
\(e + 0.15 \approx 0.19673\)
Now, we solve for \(e\):
\(e \approx 0.19673 - 0.15\)
\(e \approx 0.04673\)
Rounding to three decimal places, the calculated super elevation is approximately 0.047.
The calculated super elevation required for the given conditions is approximately 0.047.
| Parameter | Value | Units |
|---|---|---|
| Radius (R) | 100 | m |
| Design Speed (V) | 50 | km/h |
| Design Speed (V) | 13.889 (approx) | m/s |
| Friction Coefficient (f) | 0.15 | - |
| Gravity (g) | 9.81 | m/s<sup>2</sup> |
| Calculated Super Elevation (e) | 0.047 (approx) | - |
| Concept | Description | Formula |
|---|---|---|
| Super Elevation (e) | The transverse slope provided to the pavement on horizontal curves. | \(e + f = \frac{V^2}{gR}\) |
| Lateral Friction (f) | The frictional force between the vehicle's tires and the road surface resisting outward movement. | Included in the combined formula. |
| Centrifugal Force | An outward force experienced by a vehicle moving on a curve. It is balanced by super elevation and friction. | Proportional to \(V^2/R\). |
| Design Speed (V) | The maximum speed assumed for designing road features. | Used in calculations; needs unit consistency. |
Designing super elevation involves a balance. While a higher 'e' helps counteract centrifugal force, maximum limits are specified (e.g., 0.07 or 7% in plain and rolling terrain in India) to prevent slow-moving vehicles from slipping inwards, especially at low speeds or during stops. If the calculated 'e' exceeds the maximum limit, the radius of the curve might need to be increased, or the design speed might need to be reduced.
When calculating super elevation, it is standard practice to first calculate 'e' ignoring friction (using \(e = \frac{V^2}{gR}\) for a reduced design speed, typically 75% of the actual design speed) to find the equilibrium super elevation. If this value is less than the maximum allowed 'e', that equilibrium 'e' is adopted. If it's greater than the maximum 'e', the maximum 'e' is provided, and the required friction 'f' is then calculated for the full design speed using the formula \(f = \frac{V^2}{gR} - e_{\text{max}}\). This calculated 'f' must be less than the allowable friction coefficient (typically 0.15) to ensure safety. In this specific question, it was stated that the friction of 0.15 is fully developed, implying the combined formula \(e + f = \frac{V^2}{gR}\) should be used directly with the full design speed and friction value.
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